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(N/A) Let us join $AC$.In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB

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(N/A) Let us join $AC$.In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB Specifically,$BC > AB$.Since the angle opposite to the longer side is larger,we have $\angle BAC > \angle BCA$ ........... $(1)$In $\Delta ACD$,since $CD$ is the longest side of the quadrilateral $ABCD$,we have $CD > AD$ and $CD > AC$.Specifically,$CD > AD$.Since the angle opposite to the longer side is larger,we have $\angle CAD > \angle ACD$ ........... $(2)$Adding $(1)$ and $(2)$,we get:$(\angle BAC + \angle CAD) > (\angle BCA + \angle ACD)$$\Rightarrow \angle A > \angle C$Similarly,by joining $BD$ and using the same logic for $\Delta ABD$ and $\Delta BCD$,we can show that $\angle B > \angle D$.

$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$ (see Fig). Show that $\angle A > \angle C$ and $\angle B > \angle D$.

Similar Questions

$\Delta ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$ (see figure). Show that $\angle BCD$ is a right angle.

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$(i)$ $AD$ bisects $BC$
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